384 THE HANDBOOK OF PORTFOLIO MATHEMATICS
Now let us say that if tails occurs in two consecutive tosses, we are
ruined. Thus, we are trying to determine how many of those eight streams
see two consecutive tails. That number, divided by eight (the number of
permutuations) is therefore our “Probability of Ruin.”
The situation becomes more complex when we add in the concept of
multiples now. For example, in the previous example it may be that if the
first toss is heads, then two subsequent tosses of tails would not result in ruin
as the first play resulted in enough gain to avert ruin in the two subsequent
tosses of tails*.
We return now to assigning HPRs to our coin tosses at an optimalf
value of .25 andbof .6.
Note what happens as we increase the number of plays—in this case,
from two plays (i.e.,q=2) to three plays (q=3):
∀ 2 P 3 =
1. 5 × 1. 5 × 1. 5 = 3. 375
1. 5 × 1. 5 ×. 75 = 1. 6875
1. 5 ×. 75 × 1. 5 = 1. 6875
1. 5 ×. 75 ×. 75 =. 84375. 75 × 1. 5 × 1. 5 = 1. 6875
. 75 × 1. 5 ×. 75 =. 84375
. 75 ×. 75 × 1. 5 =. 84375 (ruin)
. 75 ×. 75 ×. 75 =. 421875 (ruin)
Only the last two sequences saw our stake drop to .6 or less at any time.RR(.6)= 2 / 8 =. 25
Now for four plays:∀ 2 P 4 =
1. 5 × 1. 5 × 1. 5 × 1. 5 = 5. 0625
1. 5 × 1. 5 × 1. 5 ×. 75 = 2. 53125
1. 5 × 1. 5 ×. 75 × 1. 5 = 2. 53125
1. 5 × 1. 5 ×. 75 ×. 75 = 1. 265625
1. 5 ×. 75 × 1. 5 × 1. 5 = 2. 53125
1. 5 ×. 75 × 1. 5 ×. 75 = 2. 53125
1. 5 ×. 75 ×. 75 × 1. 5 = 1. 265625
1. 5 ×. 75 ×. 75 ×. 75 =. 6328125. 75 × 1. 5 × 1. 5 × 1. 5 = 2. 53125
. 75 × 1. 5 × 1. 5 ×. 75 = 1. 265625